Dropped Rock Kinetic Energy Problem

[NIT14]

Ok, I missed last class period, but I have the general idea of this I think. I just don't know how to put it all together for this one...

A rock that weighs 40 N is dropped from a cliff 20 m above the ground. What is the kinetic energy of the rock when it gets to the ground? (disregard air friction).

Ok, I know that F=MA, I know that KE=1/2mv^2, and I know that PE=mgh.

Am I missing something here, or am I just retarded? lol. I got the an answer of 10290J and I don't even remember how!.. and I really don't think this is correct. Can anyone help? Thanks!

 

[nautica]

Are you considering an elastic collision. If not, the rock will not be moving after being stopped by the ground and if it is not moving then it has no KE.

Or is the question asking just before it hits the ground?

Reread the question and retype it.

Nautica

 

[NIT14]

I'm pretty sure it means right before it gets to the ground. If not, then the energy would be converted to heat engery, right?

I don't think it has anything to do with "elastic collision" either.

 

[nautica]

Okay, you know the acceleration is 9.8 m/s^2. Use that to find the velocity at just before it hits the ground.

Use that velocity and plug it into you KE equation.

Does that help.

Nautica

 

[ShawnD]

Originally posted by NIT14
A rock that weighs 40 N is dropped from a cliff 20 m above the ground. What is the kinetic energy of the rock when it gets to the ground? (disregard air friction).

It's sometimes best to lead by example instead of giving hints on how to do it.

There are several ways to get the answer to this question. The easiest by far is to use the work formula.

W = Fd
W = (40N)(20m)
W = 800J

The potential energy way of looking at it is exactly the same since mg is the same as F.
E = mgh
E = (40N)(20m)
E = 800J

Then there is the kinetic energy way which is just a waste of time since you have to find the velocity and the mass. It's also less accurate because you have to do more operations.

m = F/a
m = 40/9.81
m = 4.0775kg

Vf^2 = Vi^2 + 2ad
Vf^2 = 2(9.81)(20)
Vf = 19.809m/s

E = (1/2)mv^2
E = (1/2)(4.0775)(19.809)^2
E = 799.998J


Don't pay much attention to how many significant figures I use.

 

[NIT14]

Hey, thanks a lot guys! Makes perfect sense now.